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- Feb 14, 2012
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Solve for $x$ in terms of $a$ in the equation $x^4-3ax^3+3a^3x+a^4=0$.
Hi Dan,Great. Now I'm not getting any sleep tonight.
-Dan
Actually, I did.Hi Dan,
I take this as you like this challenge problem and I wish you the best of luck to crack it it no time!![]()
Thanks for participating, kaliprasad...my solution
X^4 – 3ax^3 + 3a^3 x + a^4 = 0
Forgetting the sign it is symmetric from left to right and a is increasing as x is decresing
So divide by x^2 and collect x^n and 1/x^n together
(X^2 + a^4/x^2) – 3a(x - a^2 /x) = 0
Or (x-a^2/x)^2 + 2a^2 – 3a(x-a^2/x) = 0
Putting x- a^2/x = y we get y^2 + 2a^2 – 3ay = 0
So y = a or 2a
So x^2 - 2ax – a^2 = 0 or x^2 –ax – a^2 = 0
both quadratic can be solved
I did not check you calculation.Thanks for participating, kaliprasad...
But I don't quite agree with the end result that you obtained.
Take for example if \(\displaystyle x^2 - 2ax – a^2 = 0\) is true, solving it for $x$ we get:
\(\displaystyle x=a\pm a\sqrt{2}=a(1\pm \sqrt{2})\)
Substitute the expression of $x=a(1+\sqrt{2})$ above into the original given equation, we have:
\(\displaystyle x^4 – 3ax^3 + 3a^3 x + a^4=(a(1+ \sqrt{2}))^4 – 3a(a(1+ \sqrt{2}))^3 + 3a^3(a(1+ \sqrt{2})+ a^4=-12a^4 \ne 0\)
What do you think, kaliprasad?![]()